Question

If $\tan^{-1} (x^2 - y^2) = a$, where $a$ is a constant, then $\frac{dy}{dx}$ is:

• A. $\frac{x}{y}$
• B. $-\frac{x}{y}$
• C. $\frac{a}{y}$
• D. $\frac{a}{x}$

Hint:

When differentiating an inverse trigonometric function, apply the chain rule carefully to account for both the function and its argument.

Answer 1

The Correct Option is B

We are given that $\tan^{-1}(x^2 - y^2) = a$. Differentiating both sides with respect to $x$, we get: \[ \frac{d}{dx} \left[\tan^{-1}(x^2 - y^2)\right] = \frac{d}{dx} [a]. \] Since $a$ is a constant, its derivative is zero. Now, using the chain rule for differentiation: \[ \frac{1}{1 + (x^2 - y^2)^2} \cdot \frac{d}{dx}(x^2 - y^2) = 0. \] The derivative of $x^2 - y^2$ with respect to $x$ is: \[ \frac{d}{dx}(x^2 - y^2) = 2x - 2y\frac{dy}{dx}. \] Thus, the equation becomes: \[ \frac{2x - 2y\frac{dy}{dx}}{1 + (x^2 - y^2)^2} = 0. \] For this equation to hold, we must have: \[ 2x - 2y \frac{dy}{dx} = 0. \] Solving for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = -\frac{x}{y}. \]